24 May 2014 the Lemniscate of Bernoulli, Without Formulas

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How to construct the lemniscate of Bernoulli? There exists a very simple method to construct the lemniscate of Bernoulli using the following three-bar linkage. This construction was invented by James Watt: Take two equal rods F1A and F2B of the 1 length F1F2 and ﬁxed at the points F1 and F2 respectively. Let points A and B lie on √2 | | opposite sides of the line F1F2. The third rod cinnects the points A and B and its length equals F1F2 (Fig. 3). Then, during the motion of the linkage the midpoint X of the rod | | AB traces the lemniscate of Bernoulli with foci at F1 and F2.

F1 F2

B Fig. 3.

Proof. Note that the quadrilateral F1AF2B is an isosceles trapezoid (Fig. 4). Moreover, triangles AF1X and ABF1 are similar, because they have the common angle A and △ △ the following relation on their sides holds:

AF1 AB | | = | | = √2. AX AF1 | | | | For the same reason, triangles BXF2 and BF2A are similar. They have the com- △ △ mon angle B, and the ratios of the length of the sides with endpoints at B is √2. Therefore, we can write the following equations:

∠AF1X = ∠ABF1 = ∠BAF2 = ∠XF2B. A

F1 O F2

B Fig. 4.

Let us remark that in the trapezoid F1AF2B angles ∠A and ∠F2 are equal. Since angles ∠XAF2 and ∠XF2B are equal too, we obtain ∠F1AX = ∠XF2A. This implies that triangles F1AX and AF2X are similar. Therefore, △ △ F1X AX 2 2 | | = | | XF1 XF2 = AX = F1O . AX XF2 ⇒ | | · | | | | | | | | | | 2 Thus, we have shown that point X lies on the lemniscate of Bernoulli Since the motion of the point X is continuous and X attains the farthest points of the lemniscate, the trajectory of X is the whole lemniscate of Bernoulli. A

MM OO′′′′′ OOX O F1 F2 N B Fig. 5.

Let O be the midpoint of the segment F1F2 (double point of the lemniscate). Denote by M and N the midpoints of the segments F1A and F1B respectively (Fig. 5). Translate the point O by the vector −−→NF1. Denote the new point by O′. Observe that triangles F1MO and NXO are congruent. Moreover, the following equation holds: △ ′ △ 1 F1M = F1O′ = F1O . | | | | √2| | 1 In other words, the points M and O lie on the circle with center at F1 and radius F1O . ′ √2 | | Using this observation, we can obtain another elegant method to construct the lemniscate of Bernoulli. B AA F1 OO F2 X Fig. 6. 1 Let us construct the circle with center at one of the foci and radius F1O . On each √2 | | secant OAB (where A and B are the points of intersection of the circle and the secant) chose points X and X such that AB = OX = OX (Fig. 6). The union of all points ′ | | | | | ′| X and X′ form the lemniscate of Bernoulli with foci F1 and F2. A C Y B

F1 OO F2

Fig. 7. Another interesting way to construct lemniscate with the linkages given in Figure 7. The lengths of the segments F1A and F1O are equal. The point A is the vertex of rods AX and AY of the length √2 F1O . Denote the midpoints of these rods by B and C, | | AX and join them with O by the another rod of the length | 2 | . In the process of rotating of point A around the circle each of the points X and Y generates a half of the lemniscate of Bernoulli with foci F1 and F2.

3 The lemniscate of Bernoulli and equilateral hyperbola. The hyperbola is a much more well-known curve. Hyperbola with foci F1 and F2 is a set of all points X such that value F1X F2X is constant. Points F1 and F2 are called the foci of the | | − | | hyperbola. Among all hyperbolas, we single out equilateral or rectangular hyperbolas, i. F1F2 e., the set of points X such that F1X F2X = | | . | | − | | √2 The lemniscate of Bernoulli is an inversion image of an equilateral hyperbola. Before proving this claim, let us recall the deﬁnition of a inversion. Deﬁnition 1. Inversion with respect to the circle with center O and radius r is a trans- formation which maps every point X in the plane to the point X∗ lying on the ray OX r2 such that OX∗ = OX . | | | | The inversion has many interesting properties, see, for example [2]. Among the properties, is the following: a circle or a line will invert to either a circle or a line, depending on whether it passes through the origin. We will prove here just one simple lemma that will help us later. Lemma 1. Suppose A is an orthogonal projection of the point O on some line ℓ. Then inversion image of the line ℓ with respect to a circle with center at O is the circle with diameter OA∗, where A∗ is the inversion image of the point A. A ∗∗∗ AAA ℓ B ∗ O1 B∗ O Fig. 8.

Proof. Let B be any point on the line ℓ, and let B∗ be its image (Fig. 8). Since r2 r2 OA∗ = and OB∗ = , | | OA | | OB | | | | we obtain that triangles OAB and OB A are similar. Therefore the angle ∠OB A △ △ ∗ ∗ ∗ ∗ is a right angle and the point B∗ lies on the circle with diameter OA∗.

Note that the center O1∗ of this circle is the inversion of the point O1, where O1 is the point symmetric to O with respect to the line ℓ. Now, let us prove that the lemniscate of Bernoulli with foci F1 and F2, is an inversion of the equilateral hyperbola with foci F1 and F2 with respect to the circle with center at O and radius OF1 . For this proof, we will use the results we obtained in the proof of | | correctness of the ﬁrst method to construct the lemniscate (Fig. 4). Let P be the point of the intersection of the lines F1A and F2B and let Q be symmetric point to P with respect to the line F1F2 (Fig. 9). Note that

F1F2 F2Q F1Q = F2P F1P = AP F1P = F1A = | |. | | − | | | | − | | | | − | | | | √2

Therefore, the points P and Q lie on the equilateral hyperbola with foci at F1 and F2. Now it remains to show that points X and Q are the images of each other under the inversion with center at O and radius OF1 . First, let us show that triangles F1XO | | △ and P F1O are similar. △ 4 Q A

X OO F1 F2

P Fig. 9.

The quadrilateral F1XOB is a trapezoid. Therefore ∠OXF1 + ∠XF1B = 180◦. Also, we have ∠AF1O + ∠OF1P = 180◦. Since the angle ∠XF1B is equal to the angle ∠AF1O, we obtain that ∠OXF1 and ∠OF1P are equal to each other. Since angels ∠XF2B and ∠XF1A are equal, we have that ∠XF1P + ∠P F2X = 180◦. In other words, the quadrilateral P F1XF2 is inscribed. Therefore, we have

∠F2F1X = ∠F2P X = ∠F1P O.

The last equation provided that points O and X are symmetric to each other with respect to the perpendicular bisector of the segment F1B. Thus, triangles F1XO and P F1O are similar because they have two corresponding △ △ pairs of equal angles. It follows that ∠F1OX and ∠F1OP are equal, and we have that the point Q lies on the ray OX. In addition, from similarity of triangles F1XO and △ QF1O (it is congruent to the triangle P F1O), we obtain △ △ OX OF1 2 | | = | | OX OQ = OF1 . OF1 OQ ⇒ | | · | | | | | | | | It means that points Q and X are images of each other under the inversion with center at O and radius OF1 . | |

X QQ F1 O F2 PP ℓ

Fig. 10.

If we look at Figure 9, we can make another observation: the points X and O lie on the circle with centered at P . It is interesting that this circle touches the lemniscate of Bernoulli.

Proof. Suppose ℓ is a tangent line to the hyperbola in the point Q. From Lemma 1, it follows that the image of the line ℓ under the inversion with center at O and radius F1O | |

5 is a circle ωℓ passing through the point O. Since X is the inversion image of the point Q, we see that the circle ωℓ touches the lemniscate at the point X. From the same Lemma we conclude that the center of this circle lies on the normal line from the point O to the line ℓ. Let us show that lines OP and OQ are symmetric to each other with respect to the line F1F2. It follows that the point P is the center of the circle ωℓ.

S Q

O R Fig. 11. 1 Without loss of generality, we can assume that the equation of the hyperbola is y = x . Suppose line ℓ intersects the abscissa and the ordinate in the points R and S respectively 1 (Fig. 11). It is well-known that the derivative of the function x at the point x0 is equal 1 to −2 . It follows that the point Q is the midpoint of the segment RS, and OQ is the x0 median of the right triangle ROS. Therefore, the angles ∠QOR and ∠QRO are equal. △ Since angles ∠POS and ∠QOR are also equal, we obtain that the lines OP and RS are perpendicular, as was to be proved. X 222ααα

F1 α F2 OO PPP

Fig. 12.

Let us note the following: Since the circle ωℓ touches the lemniscate at the point X, the radius P X of this circle is a normal (perpendicular to the tangent line) to the lemniscate at X (Fig. 12). Note that the triangle XP O is isosceles, and the lines XO and P O are △ symmetric with respect to the line F1O. Therefore, we can write these equations:

∠P XO = ∠XOP = 2∠P OF1. It follows that there exists the following very simple method to construct the normal to the lemniscate of Bernoulli: For any point X on the lemniscate, take the line forming with the line intersecting OX at X, with the angle of intersection equal to 2∠XOF . This line will be a normal to the lemniscate.

References

[1] http://xahlee.info/SpecialPlaneCurves dir/LemniscateOfBernoulli dir/lemniscateOfBernoulli.html. [2] R. A. Johnson. Advanced Euclidean Geometry. Dover Publications New York, N. Y, 2007. [3] J. D. Lawrence. A catalog of special plane curves. Dover Publications New York, N. Y, 1972.